Question 2 of 7: Absolute, Relative and Effective Permeability from Core-Flood Data

Question 2: Absolute, Relative and Effective Permeability from Core-Flood Data (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Relative permeability curves (read from the exam's own chart, below): $k_{ro}=0.9$, $k_{rw}=0$ at $S_w=20\%$ (connate water); $k_{ro}=0.4$, $k_{rw}=0.05$ at $S_w=40\%$; both curves cross at $k_{ro}=k_{rw}\approx0.15$ near $S_w=55\%$; $k_{ro}=0$, $k_{rw}=0.2$ at $S_w=80\%$ (residual oil).

[Figure not reproduced: Fig. Q2: oil and water relative permeability vs. water saturation, from the exam's chart (anchor points as read off the curve). See the official exam paper.]

Find. (a) absolute permeability $k$; (b) $q_o$ and $q_w$ at $S_w=40\%$; (c) $k_{w,eff}$ at residual oil saturation and $k_{o,eff}$ at connate water saturation; (d) why $k_{ro}k+k_{rw}k\lt k$ always.

Approach. Use Darcy's linear-flow equation in Darcy units (cm, cm², cP, atm, cm³/s) throughout: first back out the absolute permeability from the single-phase water flood, then scale it by the relative permeabilities read off the chart at each saturation of interest to get effective permeabilities and, from those, the two-phase rates.

1. Absolute permeability from the 100%-water flood. Darcy's law: $q=\dfrac{kA}{\mu}\dfrac{\Delta p}{L}\ \Rightarrow\ k=\dfrac{q\mu L}{A\,\Delta p}=\dfrac{0.01(1)(20)}{4(2)}$, so $\boxed{k=0.025\ \text{Darcy}=25\ \text{mD}}$.
2. Effective permeabilities at $S_w=40\%$. From the chart, $k_{ro}=0.4$ and $k_{rw}=0.05$ at $S_w=40\%$: $k_{o,eff}=k_{ro}\,k=0.4(0.025)=0.010$ Darcy, $k_{w,eff}=k_{rw}\,k=0.05(0.025)=0.00125$ Darcy.
3. Oil and water rates at $S_w=40\%$. $q_o=\dfrac{k_{o,eff}A}{\mu_o}\dfrac{\Delta p}{L}=\dfrac{0.010(4)(3)}{2(20)}$, so $\boxed{q_o=3.0\times10^{-3}\ \text{cm}^3/\text{s}}$. $q_w=\dfrac{k_{w,eff}A}{\mu_w}\dfrac{\Delta p}{L}=\dfrac{0.00125(4)(3)}{1(20)}$, so $\boxed{q_w=7.5\times10^{-4}\ \text{cm}^3/\text{s}}$.
4. Effective permeabilities at the saturation end-points. Residual oil saturation is where the oil curve reaches zero, at $S_w=80\%$, where the chart reads $k_{rw}=0.2$: $k_{w,eff}\big|_{S_{or}}=0.2(0.025)$, so $\boxed{k_{w,eff}=0.005\ \text{Darcy}=5\ \text{mD}}$. Connate water saturation is where the water curve starts from zero, at $S_w=20\%$, where the chart reads $k_{ro}=0.9$: $k_{o,eff}\big|_{S_{wc}}=0.9(0.025)$, so $\boxed{k_{o,eff}=0.0225\ \text{Darcy}=22.5\ \text{mD}}$.
5. Why $k_{o,eff}+k_{w,eff}\lt k_{abs}$ at any saturation. When two immiscible phases share the pore space, each phase's own flow channels are narrowed and made more tortuous by the presence of the other phase (film coatings on grain surfaces, blocked pore throats, snap-off of the non-wetting phase into isolated blobs), and part of the pore network carries no flow of either phase at all (e.g., pores fully occupied by the immobile/trapped phase). Both effects reduce the cross-sectional area and increase the flow-path length available to each phase relative to single-phase flow, so $k_{ro}+k_{rw}\lt 1$ and hence $k_{o,eff}+k_{w,eff}\lt k_{abs}$ over the whole two-phase saturation range.